We want to know what degree this curve is.
|This was the picture originally shown in La géométrie. Choose an arbitrary straight line KA and fix the ruler (green) at G and place it at an arbitrary point L.|
The same equation will result using any point L. To see a slide show of this, click on the link below. You need ghostview to open this.
|First, fix the ruler at the point L. Then construct a line parallel to GA at the point C. This creates the line CB with a length of y.|
|Now, connect C and K and construct NL to be parallel to CB and GA.|
|Zooming in on triangle CBK, we can see that triangle CBK is similar to triangle NLK since all three angles are equal. Thus, all the sides are proportional. So c/b = y/BK, making BK = yb/c|
To see a proof on similar triangles, click on the link below. This proof is by David Hilbert (1862-1943).
Again, you need ghostview to open it.
|Therefore, BL = yb/c - b|
BK = yb/c
BL = yb/c - b
AK = x + yb/c - b
Triangle GAL is similar to triangle CBL since all three angles are equal.
|y/(yb/c - b) = a/(x +
yb/c - b)
y(x + yb/c - b) = a(yb/c - b)
yx + y2b/c - yb = yab/c -ab
y2b/c = yab/c - ab + yb - yx
y2 = ya - ac + yc - yxc/b
Since the highest degree is two, this curve is a conic section. In fact, it is a hyperbola.
Now that we know it's a hyperbola, the next problem that we want to address are the asymptotes.
Descartes ended his proof here without addressing this problem.
This next construction and proof is by Van Schooten.
|From the slide show, we saw that EA = NL. Thus, EA = c and GA = a.|
Extend AG to make a point D with DG = EA.
Then DG = NL = c.
Draw DF parallel to KC.
Claim: DF and AF are the asymptotes to the hyperbola GCE.
|Proof: Extend BC to intersect DF at I.|
Draw DH parallel to AF intersecting BC at H.
Triangle NKL is similar to triangle IDH since all three angles are equal.
Thus, KL/LN = DH/HI
DH = AB = x, KL = b, and LN = c.
Therefore b/c = x/HI, so HI = cx/b
Since HB = DA, HB = c + a
Therefore IB = a + c - cx/b and IC = a + c - cx/b - y
(IC)(BC) = (DE)(EA), a property of a hyperbola with respect to the distances between it and its asymptotes.
Thus, (a + c - cx/b - y)y = ac
or y2 = cy - cxy/b + ay - ac, which is the same equation as before.
Therefore, AF and DF are the asymptotes to the hyperbola GCE.